Question

∫dx/x2+x+1 and ∫dx/(x^2 -x) are my questions. Thanks for helping.

Answer 1

lajeik? is that u

Answer 2

lajeik fleming

Answer 3

Lajeik Fleming? the little black sleek-feathered one

Answer 4

omfg its lajeik

Answer 5

Have you tried partial fractions on this one?

Answer 6

\[(x^2 - x + 2) / ((x-1)(x^2 + x + 1) ) = A/(x+1) + (Bx + C)/(x^2 + x +1) \]

No I haven't. Is this how I begin?

Answer 7

Yes, your decomposition so far is good. All that's left is finding A, B, and C.

Answer 8

ok thank you I have found A B and C which leads me to integrate \[(x-4)/(x^2 + x + 1)\] How do I integrate this?

Answer 9

So, you're integrating this
\[\frac{x-4}{x^2+x+1}\]
or
\[\frac{x^2-x+2}{(x-1)(x^2+x+1)}?\]
Sorry, I'm a bit confused.

Answer 10

after partial fractions I am left with a longer integral of \[2/3 \int\limits_{} 1/(x+1) dx +1/3 \int\limits_{} (x-4)/(x^2 + x + 1) dx\] How do I start the second integral ?

Answer 11

should be 1/(x-1) in the first part..

Answer 12

Ah, thanks.
\[\int\frac{x-4}{x^2+x+1}dx\]
Notice that the derivative of the denominator is 2x + 1. Rewriting the numerator, you get
\[\frac{1}{2}\int\frac{2x-8}{x^2+x+1}dx\\
\frac{1}{2}\int\frac{2x+1-9}{x^2+x+1}dx\\
\frac{1}{2}\left[\int\frac{2x+1}{x^2+x+1}dx-\frac{1}{2}\int\frac{9}{x^2+x+1}dx\right]
\]

Answer 13

Sorry, the last line should be
\[\frac{1}{2}\left[\int\frac{2x+1}{x^2+x+1}dx-\int\frac{9}{x^2+x+1}dx\right]\]

Answer 14

Now, making the substitution for the first integral:
\[u=x^2+x+1\\
du=(2x+1)\;dx\]
\[\frac{1}{2}\left[\int\frac{1}{u}du-\int\frac{9}{x^2+x+1}dx\right]\]
For the second integral, complete the square in the denominator.

Answer 15

Okay. We haven't learned this in class and I don't think it will be on the test. I thank you so much for explaining this much so far. What math is this taught in? I'm in Calc 272 at a community college and I'm wondering when I have to know this. lol.. Thanks so much for your help.. wow.. unbelievable I haven't been taught this yet.

Answer 16

There isn't really an explicit name for this method. Just some algebraic manipulation to get what you want.

Answer 17

What math did you learn this in?

Answer 18

BC calc, two years ago

Answer 19

Thank you so much for your help!

Answer 20

You're welcome. If you're curious, the last integral (with the completing the square) later uses trig substitution. If you've learned that, it might be good practice.

Answer 21

Sure I'll try it.

Answer 22

∫dx/x2+x+1

Answer 23

∫dx/(x^2 -x)

Answer 24

∫dx/x2+x+1 and ∫dx/(x^2 -x) are my questions. Thanks for helping.